+9466 Let y = x4 - 2x2
\(y\,=\,x^4-2x^2\\~\\ \frac{dy}{dx}\,=\,\frac{d}{dx}x^4-\frac{d}{dx}2x^2\\~\\ \frac{dy}{dx}\,=\,4x^3-4x\)
Now let's find what values of x make \(\frac{dy}{dx}\) be 0
\(0\,=\,4x^3-4x\\~\\ 0\,=\,4x(x^2-1) \)
Set each factor equal to zero and solve for x
\(\begin{array} {lcl} 4x=0&\qquad\text{or}\qquad &x^2-1=0\\~\\ x=0&\text{or}&x^2=1\\~\\ &&x=1\quad\text{or}\quad x=-1 \end{array}\)
These are the x values of all the critical points.
We can look at a graph to determine which is the minimum:
https://www.desmos.com/calculator/vvl7egtes1
We can see that the minumum occurs when x = 1 and when x = -1
When x = ±1 , x2 = 1
And when x2 = 1 ,
y = x4 - 2x2 = (x2)2 - 2x2 = (1)2 - 2(1) = -1
So the minumum value of y is -1
+9466 Let y = x4 - 2x2
\(y\,=\,x^4-2x^2\\~\\ \frac{dy}{dx}\,=\,\frac{d}{dx}x^4-\frac{d}{dx}2x^2\\~\\ \frac{dy}{dx}\,=\,4x^3-4x\)
Now let's find what values of x make \(\frac{dy}{dx}\) be 0
\(0\,=\,4x^3-4x\\~\\ 0\,=\,4x(x^2-1) \)
Set each factor equal to zero and solve for x
\(\begin{array} {lcl} 4x=0&\qquad\text{or}\qquad &x^2-1=0\\~\\ x=0&\text{or}&x^2=1\\~\\ &&x=1\quad\text{or}\quad x=-1 \end{array}\)
These are the x values of all the critical points.
We can look at a graph to determine which is the minimum:
https://www.desmos.com/calculator/vvl7egtes1
We can see that the minumum occurs when x = 1 and when x = -1
When x = ±1 , x2 = 1
And when x2 = 1 ,
y = x4 - 2x2 = (x2)2 - 2x2 = (1)2 - 2(1) = -1
So the minumum value of y is -1
